YES

We show the termination of the TRS R:

  a__f(X) -> a__if(mark(X),c(),f(true()))
  a__if(true(),X,Y) -> mark(X)
  a__if(false(),X,Y) -> mark(Y)
  mark(f(X)) -> a__f(mark(X))
  mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
  mark(c()) -> c()
  mark(true()) -> true()
  mark(false()) -> false()
  a__f(X) -> f(X)
  a__if(X1,X2,X3) -> if(X1,X2,X3)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: a__f#(X) -> mark#(X)
p3: a__if#(true(),X,Y) -> mark#(X)
p4: a__if#(false(),X,Y) -> mark#(Y)
p5: mark#(f(X)) -> a__f#(mark(X))
p6: mark#(f(X)) -> mark#(X)
p7: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)
p8: mark#(if(X1,X2,X3)) -> mark#(X1)
p9: mark#(if(X1,X2,X3)) -> mark#(X2)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: a__if#(false(),X,Y) -> mark#(Y)
p3: mark#(if(X1,X2,X3)) -> mark#(X2)
p4: mark#(if(X1,X2,X3)) -> mark#(X1)
p5: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)
p6: a__if#(true(),X,Y) -> mark#(X)
p7: mark#(f(X)) -> mark#(X)
p8: mark#(f(X)) -> a__f#(mark(X))
p9: a__f#(X) -> mark#(X)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__f#_A(x1) = ((0,1),(0,0)) x1 + (1,0)
        a__if#_A(x1,x2,x3) = ((0,1),(0,0)) x2 + ((0,1),(0,0)) x3
        mark_A(x1) = x1 + (2,0)
        c_A() = (1,0)
        f_A(x1) = x1 + (0,1)
        true_A() = (1,0)
        false_A() = (1,1)
        mark#_A(x1) = ((0,1),(0,0)) x1
        if_A(x1,x2,x3) = x1 + x2 + ((0,0),(1,1)) x3 + (1,0)
        a__f_A(x1) = x1 + (2,1)
        a__if_A(x1,x2,x3) = x1 + x2 + ((0,0),(1,1)) x3 + (2,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__f#_A(x1) = (0,0)
        a__if#_A(x1,x2,x3) = (0,0)
        mark_A(x1) = (2,1)
        c_A() = (3,2)
        f_A(x1) = (0,0)
        true_A() = (3,2)
        false_A() = (1,0)
        mark#_A(x1) = (0,0)
        if_A(x1,x2,x3) = (3,0)
        a__f_A(x1) = (2,1)
        a__if_A(x1,x2,x3) = (2,1)
    

The next rules are strictly ordered:

  p7, p9

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: a__if#(false(),X,Y) -> mark#(Y)
p3: mark#(if(X1,X2,X3)) -> mark#(X2)
p4: mark#(if(X1,X2,X3)) -> mark#(X1)
p5: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)
p6: a__if#(true(),X,Y) -> mark#(X)
p7: mark#(f(X)) -> a__f#(mark(X))

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: a__if#(true(),X,Y) -> mark#(X)
p3: mark#(f(X)) -> a__f#(mark(X))
p4: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)
p5: a__if#(false(),X,Y) -> mark#(Y)
p6: mark#(if(X1,X2,X3)) -> mark#(X1)
p7: mark#(if(X1,X2,X3)) -> mark#(X2)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__f#_A(x1) = ((1,1),(0,0)) x1 + (1,0)
        a__if#_A(x1,x2,x3) = ((0,1),(0,0)) x1 + ((1,1),(1,0)) x2 + ((1,1),(1,0)) x3
        mark_A(x1) = ((1,1),(0,1)) x1
        c_A() = (0,0)
        f_A(x1) = ((1,1),(0,1)) x1 + (0,1)
        true_A() = (0,0)
        mark#_A(x1) = ((1,1),(1,0)) x1
        if_A(x1,x2,x3) = x1 + ((1,1),(0,1)) x2 + ((1,0),(1,1)) x3
        false_A() = (1,1)
        a__f_A(x1) = ((1,1),(0,1)) x1 + (1,1)
        a__if_A(x1,x2,x3) = x1 + ((1,1),(0,1)) x2 + ((1,1),(1,1)) x3
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__f#_A(x1) = (9,0)
        a__if#_A(x1,x2,x3) = x2 + ((0,1),(0,0)) x3 + (1,0)
        mark_A(x1) = ((1,1),(0,1)) x1 + (1,0)
        c_A() = (1,0)
        f_A(x1) = ((0,1),(1,1)) x1 + (14,4)
        true_A() = (1,1)
        mark#_A(x1) = x1
        if_A(x1,x2,x3) = x1 + ((1,1),(0,1)) x2 + ((0,1),(0,0)) x3 + (3,3)
        false_A() = (1,1)
        a__f_A(x1) = ((1,1),(0,1)) x1 + (13,3)
        a__if_A(x1,x2,x3) = x1 + ((1,1),(0,1)) x2 + ((0,1),(0,0)) x3 + (4,3)
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7

We remove them from the problem.  Then no dependency pair remains.